Question

JEE Mains - 2019

Question # 2: Refer to the attachment ^^

Kindly answer with a full explanation.​

Answer 1

Answer :-

$\mathsf{M^{-1}L^{-2}T^3A^2}$

Option → D

Given :-

$\mathsf{\sqrt{\dfrac{\varepsilon_0}{\mu_0}}}$

To find :-

It's dimensions.

Solution:-

• From coulumb's law.

→$\mathsf{F = K \dfrac{q_1.q_2}{r^2}}$

• The dimension of charge is $\mathsf{ Q = AT }$

→$\mathsf{K = \dfrac{F \times r^2}{q_1.q_2}}$

→$\mathsf{K = \dfrac{F \times r^2}{q^2 }}$

→$\mathsf{K = \dfrac{MLT^{-2}\times L^2}{(AT)^2}}$

→$\mathsf{ K = \dfrac{ML^3T^{-2}}{A^2T^2}}$

→$\mathsf{ K = ML^3 T^{-4} A^{-2}}$

→$\mathsf{K = \dfrac{1}{4 \pi \varepsilon_0}}$

• Numbers are dimensionless.

→$\mathsf{\varepsilon_0 = \dfrac{1}{K}}$

→$\mathsf{ \varepsilon_0 = \dfrac{1}{ML^3T^{-4}A^{-2}}}$

→$\mathsf{\varepsilon_0 = M^{-1}L^{-3}T^{4}A^{2}}$

• From magnetic permiability formula.

→$\mathsf{C^2 = \dfrac{1}{\mu_0 \varepsilon_0}}$

• C is the speed of light.

• $\mathsf{ \epsilon_0}$ is the permiability of free space.

→$\mathsf{\mu_0 = \dfrac{1}{\varepsilon_0 C^2 }}$

→$\mathsf{\mu_0 = \dfrac{1}{M^{-1}L^{-3}T^{4}A^{2} \times L^2 T^{-2}}}$

→$\mathsf{ \mu_0 = \dfrac{1}{M^{-1} L^{-1}T^2A^2}}$

→$\mathsf{ \mu_0 = M^1L^{1}T^{-2}A^{-2}}$

→$\mathsf{\sqrt{\dfrac{\varepsilon_0}{\mu_0}}}$

• use the above dimensional formula.

→$\mathsf{\sqrt{\dfrac{M^{-1}L^{-3}T^{4}A^{2}}{M^{1}L^{1}T^{-2}A^{-2}}}}$

→$\mathsf{\sqrt{M^{-1-1}L^{-3-1}T^{4+2}A^{2+2}}}$

→$\mathsf{\sqrt{M^{-2}L^{-4}T^{6}A^{4}}}$

→$\mathsf{M^{-1}L^{-2}T^3A^2}$

hence,

Required Dimension formula is $\mathsf{M^{-1}L^{-2}T^3A^2 }$